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# find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. assume that the items are identical in shape and size.

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## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random.

From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random ... X (b) Mean of X (c) Variance of X ## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random.

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## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

Click here👆to get an answer to your question ✍️ From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X. Question

## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

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Updated on : 2022-09-05

Solution Verified by Toppr

We have a lot of 6 items, 2 are defective →6−2=4 are non-defective.

P (drawing a defective item) =

6 2 ​ = 3 1 ​

P (drawing a non-defective item) = 1 - P (drawing a defective item) =1−

3 1 ​ = 3 2 ​

Note:4 items can be drawn from a lot of 6 items in

6 C 4 ​ ways.

Let X be the random variable of drawing 2 defective items from the lot.

P(X=0)= P (no defective item in the sample) =

6 C 4 ​ 4 C 4 ​ ​ = 15 1 ​

P(X=1)= P (one defective item from the lot) = P(one defective, 3 non-defective items) =(

6 C 4 ​ 2 C 1 ​ 4 C 3 ​ ​ )= 15 8 ​

P(X=2)= P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =

6 C 4 ​ 2 C 2 ​ 4 C 2 ​ ​ = 15 6 ​

Therefore the required probability distribution is as follows:

X:         0    1    2

P(X): 15 1 ​ 15 8 ​ 15 6 ​

So Variance of X will be :E(X

2 )−(E(X)) 2 E(X)=∑ i=0 2 ​ x i ​ P(X=x i ​ ) →0× 15 1 ​ +1× 15 8 ​ +2× 15 6 ​ = 15 20 ​ E(X 2 )=∑ i=0 2 ​ x i 2 ​ P(X=x i ​ ) →0 2 × 15 1 ​ +1 2 × 15 8 ​ +2 2 × 15 6 ​ = 15 32 ​ Variance of X= 15 32 ​ −( 15 20 ​ ) 2 = 45 16 ​ Video Explanation Solve any question of Probability with:-

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## [Sample Paper] Find the mean number of defective items in a sample of

Question 27 (Choice 2) Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size. Let X : Number of defective items drawn Since ther Check sibling questions

## Question 27 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 19, 2022 by Teachoo

## Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size. This is a modal window.

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Question 27 (Choice 2) Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size. Let X : Number of defective items drawn Since there are maximum of 2 defective items We can get 0, 1 or 2 defective items So, X = 0 or X = 1 or X = 2 Finding probabilities separately For X = 0 Two items drawn, without replacement, 0 are defective P(X = 0) = P(not defective) × P(not defective) = 4/6 × 3/5 = 𝟐/𝟓 For X = 1 Two items drawn, without replacement, 1 is defective P(X = 1) = P(not defective) × P(defective) + P(defective) × P(not defective) = 𝟒/𝟔 × 𝟐/𝟓 + 𝟐/𝟔 × 𝟒/𝟓 = 8/30+8/30 = 16/30 = 𝟖/𝟏𝟓 For X = 2 Two items drawn, without replacement, 2 are defective P(X = 2) = P(defective) × P(defective) = 2/6 × 1/5 = 𝟏/𝟏𝟓 So, probability distribution is Now, We need to find Mean or Expectation of X Mean or Expectation of X = E(X) = ∑_(𝒊 = 𝟏)^𝒏▒𝒙𝒊𝒑𝒊 = 0 ×2/5+1 × 8/15+2 × 1/15 = 8/15+ 2/15 = 10/15 = 𝟐/𝟑

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